POJ2513 【并查集+欧拉路径+trie树】

题目链接：http://poj.org/problem?id=2513

Colored Sticks

Time Limit: 5000MS		Memory Limit: 128000K
Total Submissions:40949		Accepted: 10611

Description

You are given a bunch of wooden sticks. Each endpoint of each stick is colored with some color. Is it possible to align the sticks in a straight line such that the colors of the endpoints that touch are of the same color?

Input

Input is a sequence of lines, each line contains two words, separated by spaces, giving the colors of the endpoints of one stick. A word is a sequence of lowercase letters no longer than 10 characters. There is no more than 250000 sticks.

Output

If the sticks can be aligned in the desired way, output a single line saying Possible, otherwise output Impossible.

 

题目大意：给出以下字符串，代表每根棍子的两端颜色，只能将颜色相同的端点连接起来。问是否能连成一欧拉路径。

思路：

1.棍子两端无方向要求，故是无向图的欧拉路径，无向图判断欧拉路径的充要条件是：图连通（无向图的连通性可以用并查集来判断，但有向图不可以）+ 点的度全为偶/当且仅当只有2个点的度为奇

2.关键在于将颜色字符串映射成整型编号，有两种思路：用map直接映射（TLE），用trie树给字符串上编号。

3.具体实现在代码中，值得注意的一点是 G++才能AC。

#include<stdio.h>
#include<string.h>
#define mem(a, b) memset(a, b, sizeof(a))
const int MAXN = 250000 * 2 + 10;//最多MAXN种颜色 即最多MAXN种不同的点 

char s1[12], s2[12];
int deg[MAXN];
int trie[MAXN][27], cnt, tot;
int id[MAXN], pre[MAXN];

int insert(char s[12])
{
    int flag = 1;
    int len = strlen(s);
    int root = 0;
    for(int i = 0; i < len; i ++)
    {
        int id = s[i] - 'a';
        if(!trie[root][id])
        {
            flag = 0;  //单词之前没出现过 
            trie[root][id] = ++ cnt;
        }
        root = trie[root][id];
    }
    if(!flag)
    {
        tot ++;
        id[root] = tot;
        return id[root];
    }
    else
        return id[root];
}

int find(int x)
{
    if(pre[x] == x)
        return x;
    else
    {
        int root = find(pre[x]);
        pre[x] = root;
        return pre[x];
    }
}

int main()
{
    for(int i = 1; i <= MAXN + 1; i ++)
        pre[i] = i;
    while(scanf("%s%s", s1, s2) != EOF)
    {
        int a = insert(s1), b = insert(s2);//返回的是颜色所对应的序号
        deg[a] ++, deg[b] ++; //点的度数 判断是否存在欧拉路径 
        int x = find(a), y = find(b); //查找根节点 
        if(x != y)
            pre[y] = x;
    }
    int flag = 1;
    for(int i = 1; i < tot; i ++) //总共有tot个不同的点 
        if(find(i) != find(i + 1))
        {
            flag = 0;
            break;
        }
    if(flag == 0)
        printf("Impossible\n");
    else
    {
        int xx = 0;
        for(int i = 1; i <= tot; i ++)
            if(deg[i] % 2)
                xx ++;
        if(xx == 0 || xx == 2)
            printf("Possible\n");
        else
            printf("Impossible\n");
    }
    return 0;
}